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Find the area of the region $ {(x,y):x^2+y^2 \leq 1 \leq x + y}.$

1 Answer

  • If we are given two curves represented by y=f(x);y=g(x),where $f(x)\geq g(x)$ in [a,b],the points of intersection of two curves are given by x=a and x=b,by taking common values of y from the equation of the two curves.
Step 1:
Given : $R=\{(x,y) : x^2+y^2\leq 1\leq x+y\}$
$\Rightarrow R=R_1\cap R_2$
$\Rightarrow R_1=\{(x,y):(x^2+y^2\leq 1\}$
$\Rightarrow R_2=\{(x,y): 1\leq x+y\}$
$x^2+y^2\leq 1$ represents the equation of the circle with centre (0,0) and radius 1.
$x+y=1$ is the equation of a straight line .
Now let us find the points of intersection of these two curve.
$\Rightarrow y=1-x$
Hence $x^2+(1-x)^2=1$
$\Rightarrow x^2+1-2x+x^2=1$
$x=0$ or $x=1$
Step 2:
Area of the required region is shaded region shown in the figure.
Step 3:
So the required area is given by
Step 4:
On integrating we get,
On applying the limits we get,
But $\sin^{-1}(1)=\large\frac{\pi}{2}$
answered Oct 4, 2013 by sreemathi.v

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